Solving exponential (of trigonometric functions) equation

In summary, the conversation discusses a difficult problem involving the expression $(2+ \sqrt{2})^{(\sin x)^2}-(2- \sqrt{2})^{(\cos x)^2}$ and how to solve it. The speaker has attempted a substitution method but is unsure if it is the correct approach. They are seeking insights from the forum and suggest using variables such as $a, b$ to simplify the problem.
  • #1
anemone
Gold Member
MHB
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Hi MHB,

Solve $(2+ \sqrt{2})^{(\sin x)^2}-(2- \sqrt{2})^{(\cos x)^2}=\left( 1+ \dfrac{1}{\sqrt{2}} \right)^{\cos 2x} -(2-\sqrt{2})^{\cos 2x}$.

This problem vexes me much because the only way that I could think of to solve this problem would be by substituting $(\sin x)^2=u$, and from there, I gotten another ugly equation $\left( 2^u-(2-\sqrt{2})=\dfrac{(2-\sqrt{2})^u(\sqrt{2}+1)^{1-2u}}{\sqrt{2}^{1-2u}}-\dfrac{2-\sqrt{2}}{(2-\sqrt{2})^u} \right)$, of which I don't think I am heading in the right direction...

I normally would employ another method to approach the problem since the first trial failed me, but the thing is, other than trying to solve this problem by the substitution skill, I don't think I can come up with another idea of how to crack it. Hence, I brought it up here with the hope to get some insights from the forum, and if any of you could provide me with any ideas of how to even start to work with this problem, I would greatly appreciate it.:)
 
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  • #2
anemone said:
Hi MHB,

Solve $(2+ \sqrt{2})^{(\sin x)^2}-(2- \sqrt{2})^{(\cos x)^2}=\left( 1+ \dfrac{1}{\sqrt{2}} \right)^{\cos 2x} -(2-\sqrt{2})^{\cos 2x}$.

Hello.

I don't know if you will serve:

[tex]a=2+\sqrt{2}[/tex]

[tex]1+\dfrac{1}{\sqrt{2}}=\dfrac{a}{2}[/tex]

[tex]2-\sqrt{2}=\dfrac{2}{a}[/tex]

[tex]\cos^2x=b[/tex]

[tex]\sin^2x=1-b[/tex]

[tex]\cos(2x)=2b-1[/tex]

Regards.
 

FAQ: Solving exponential (of trigonometric functions) equation

How do I solve exponential equations?

To solve an exponential equation, you need to isolate the variable in the base and use logarithms to solve for the variable. If the variable is in the exponent, use logarithms with the same base as the variable's base. If the variable is in the base, use logarithms with any base.

What is the difference between exponential and trigonometric equations?

An exponential equation involves a variable in the exponent, while a trigonometric equation involves trigonometric functions like sine, cosine, and tangent. Exponential equations can be solved using logarithms, while trigonometric equations require the use of trigonometric identities and algebraic manipulation.

Can exponential and trigonometric equations be solved in the same way?

No, exponential and trigonometric equations require different methods to solve. Exponential equations can be solved using logarithms, while trigonometric equations require the use of trigonometric identities and algebraic manipulation.

How do I know when to use logarithms to solve an exponential equation?

If the variable is in the exponent of an exponential equation, you will need to use logarithms to solve for the variable. If the variable is in the base, logarithms may also be used, but other methods may be more efficient.

Are there any restrictions when solving exponential or trigonometric equations?

Yes, there may be restrictions when solving exponential or trigonometric equations. For exponential equations, the base must be a positive number and cannot be equal to 1. For trigonometric equations, the values of the trigonometric functions must fall within certain ranges depending on the function (e.g. sine and cosine have a range of -1 to 1).

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